Factorisation of Polynomials: Factoring Methods, and Examples

A polynomial can be expressed as the sum of components with degrees less than or equal to the original polynomial. Factorisation of polynomials is the process of factoring. We are also aware that all monomials, binomials, trinomials, and all other expressions having any number of finite terms are called polynomials, provided no term contains a negative exponent or any variable in the denominator.

The integers multiplied to generate an original number are known as factors. For example, factors for 15 will include 3, 5, 15. This article will explore and learn to find the factors of algebraic expressions and polynomials.

Factors of Algebraic Expressions

An algebraic expression is a combination of variables and constants connected by signs of fundamental operations, i.e., \( + ,\, – ,\, \times \) and \( \div \). A product is the result of the multiplication of two or more constants or literals or both.

For example, \(8xyz\) is the product of \(8,\,x,\,y\) and \(z\).

Each constant and each literal multiplied together to form a product is called a factor of that product.

Any factor or group of factors of a product is known as the coefficient of the remaining factors. In the product of \(8xyz,\,8\), is the coefficient of \(xyz,\,8x\), is the coefficient of \(yz,\,8xy\) is the coefficient of \(z\), and so on.

If a factor is a numerical quantity, it is called a numerical coefficient of the remaining factors. If the factor involves letters, it is called a literal coefficient of the remaining factors. Consider another example.

Let’s take the term \(5a\), and the factors are \(5\) and \(a\). Note that \(5\) and \(a\) are called the irreducible factors of the term \(5a\), exactly similar to the idea of prime factors in the case of numbers.

Factorisation of Polynomials

Let us look at the example first. For the term \(6xy\), the factors are \(6,\,x\) and \(y\). The other combinations of factor pairs for the algebraic term \(6xy\) are shown below.

\(6xy = 6 \times xy\)
\(6xy = 2x \times 3y\)
\(6xy = 2 \times 3 \times x \times y\), etc

As we can see that, the above example has only one term. How can we find the factors of an algebraic expression having multiple terms? There are several methods to find the factors. The process of finding the factors of an algebraic expression is called factorisation.

Therefore, the process of writing an algebraic expression as a product of its factors is called factorisation.

Methods of Factorisation of Polynomials

A polynomial can be factorised using various methods. Below are the mentioned ways to factorise a polynomial.

1. Factorisation using common factors
2. Factorisation by regrouping
3. Factorisation using identities
4. Factorisation by splitting the middle term
5. Sum or difference of two cubes

Let us study every method in detail.

Common Factors

If the different terms of the polynomial have common factors, then the given polynomial can be factorised by the following procedure.

1. Find the HCF of all the terms of the given polynomial.
2. Divide each term of the given polynomial by HCF. Enclose the quotient within the brackets and keep the common factors outside the bracket.

The first and the foremost step of factorising a polynomial is through finding or extracting the common factors.

Lets us check out a couple of examples.

Example 1: Factorise \(15x + 20y\).

Solution: Identify the common factors. Next, write each term as a product of the common factor and its remaining factors. Then, factor out the common term.

\(15x + 20y = 3 \times 5 \times x + 2 \times 2 \times 5 \times y\)
\( = 5(3x) + 5(4y)\)
\( = 5(3x + 4y)\)

Since the last terms \(3x\) and \(4y\) do not have common factors, the factorisation is complete.

Example 2: Factorise the algebraic expression \(24ax – 40ay + 8a\).

Solution: \(24ax – 40ay + 8a = (8 \times 3 \times a \times x) – (8 \times 5 \times a \times y) + (8 \times a)\)

\( = 8a(3x) – 8a(5y) + 8a\)

\( = 8a(3x – 5y + 1)\)

Example 3: Factorise the polynomial \(10{(p – 2q)^3} + 6{(p – 2q)^2} – 20(p – 2q)\).

Solution: HCF  of the terms \(10{(p – 2q)^3},\,6{(p – 2q)^2}\) and \(20(p – 2q)\) is \(2(p – 2q)\).

Therefore, \(10{(p – 2q)^3} + 6{(p – 2q)^2} – 20(p – 2q) = 2(p – 2q)\left[ {5{{(p – 2q)}^2} + 3(p – 2q) – 10} \right]\).

Factorisation Using Regrouping

In an algebraic expression, all the terms collectively may not have a common factor, and these terms can be divided into groups such that the terms in each group have a common factor. Then we can extract the common factor to factorise the given expression. This method of factorisation is called factorisation by grouping terms.

The following procedure can factorise the given polynomial.

1. Arrange the terms of the given polynomial in groups so that each group has a common factor.
2. Factorise each group.
3. Take out the factor which is common to each group.

Let us understand the concept of factorisation using regrouping with the help of a couple of examples.

Example 1: Factorise the expression \(49a + 42c – 7ay – 6cy\).

Solution: Regroup the terms to get common factors between pairs of terms.

So, \(49a + 42c – 7ay – 6cy = (49a + 42c) – (7ay + 6cy)\)

\( = 7(7a + 6c) – y(7a + 6c)\)

\( = (7 – y)(7a + 6c)\)

Example 2: Factorise the expression \(5xy – 5x + 6 – 6y\).

Solution: \(5xy – 5x + 6 – 6y = (5xy – 5x) + (6 – 6y)\)

\( = 5x(y – 1) + 6(1 – y)\)

\( = 5x(y – 1) – 6(y – 1)\)

\( = (5x – 6)(y – 1)\)

Example 3: Factorise the following \({a^2}{x^2} + \left( {a{x^2} + 1} \right)x + a\).

Solution: \({a^2}{x^2} + \left( {a{x^2} + 1} \right)x + a = {a^2}{x^2} + a{x^3} + x + a\)

\({a^2}{x^2} + a{x^3} + x + a = \left( {{a^2}{x^2} + a{x^3}} \right) + x + a\)

\( = a{x^2}(x + a) + 1(x + a) = (x + a)\left( {a{x^2} + 1} \right)\)

Factorisation Using Identities

Some identities involve products of specific kinds of algebraic expressions. Thus we have four special products of algebraic expressions, also called standard identities, which are as follows;

1. \({(a + b)^2} = {a^2} + 2ab + {b^2}\)
2. \({(a – b)^2} = {a^2} – 2ab + {b^2}\)
3. \((a – b)(a + b) = {a^2} – {b^2}\)
4. \((x + a)(x + b) = {x^2} + (a + b)x + ab\)

As factorisation is the reverse process of expansion, we can have the following factorisation.
\({a^2} + 2ab + {b^2} = {(a + b)^2}\)
\({a^2} – 2ab + {b^2} = {(a – b)^2}\)
\({a^2} – {b^2} = (a – b)(a + b)\)
\({x^2} + (a + b)x + ab = (x + a)(x + b)\)

When an expression can be written in one of the forms on the left-hand side of the equations given above, we can factorise it accordingly. The expressions for \({(a + b)^2}\) and \({(a – b)^2}\) are called perfect square expressions.

Let’s solve some examples based on the identities.

Example 1: Factorise the polynomial \(4 + 12xy + 9{x^2}{y^2}\).

Solution: In the expression \(4 + 12xy + 9{x^2}{y^2}]\) we can observe that \(4 = {2^2}\) and \(9{x^2}{y^2} = {(3xy)^2}\). We will see whether the middle term \(12xy\) can be expressed as \(2(2)(3xy)\). If so, we can apply the perfect square factorisation.

\(4 + 12xy + 9{x^2}{y^2} = {2^2} + 2(2)(3xy) + {(3xy)^2}\)
\( = {(2 + 3xy)^2}\)

Example 2: Factorise the polynomial \(49{t^2} – 14t + 1\)

Solution: \(49{t^2} – 14t + 1 = {(7t)^2} – 2(7t)(1) + {(1)^2}\)
\( = {(7t – 1)^2}\)

Example 3: Factorise the polynomial \(49{x^2} – 9{y^2}\)

Solution: \(49{x^2} – 9{y^2} = {(7x)^2} – {(3y)^2}\)
\( = (7x + 3y)(7x – 3y)\)

Example 4: Factorise the polynomial \(3{y^2} – 75\)

Solution: \(3{y^2} – 75\)
When there is a common factor, we should factor it out first.
Therefore, \(3{y^2} – 75 = 3\left( {{y^2} – 25} \right)\)
\(3\left( {{y^2} – 25} \right) = 3\left[ {{{(y)}^2} – {{(5)}^2}} \right]\)
\( = 3(y + 5)(y – 5)\)

Factorisation by Splitting the Middle Term

We know that the expansion of the product of two linear expressions is a quadratic expression. For example,
\((2x + 7)(3x – 5) = 2x(3x – 5) + 7(3x – 5)\)
\( = 6{x^2} – 10x + 21x – 35\)
\( = 6{x^2} + 11x – 35\)

The factorisation is the reverse process of expansion. By reversing the process above, we factorise the quadratic expression into a product of two linear factors. Thus, we say, the factorized form of \(6{x^2} + 11x – 35\) is \((2x + 7)(3x – 5)\).

Now, look at the above-solved example carefully, with particular attention to the second and third lines. To factorise the expression \(6{x^2} + 11x – 35\), which is a trinomial, we split the middle term \(11x\) into two terms: \( – 10x\) and \(21x\).

This process can be referred to as factorisation of a trinomial by splitting the middle term, such that common terms can be then grouped.

For example, factorise \({x^2} – 8x + 12\).

The middle term is \( – 8x\), we have to split in such a way that the obtained numbers when multiplied together gives \(12{x^2}\).

Thus \( – 8x\) can be written as \( – 2x – 6x\).
Therefore, \({x^2} – 8x + 12 = {x^2} – 2x – 6x + 12\)
\( = x(x – 2) – 6(x – 2)\)
\( = (x – 6)(x – 2)\)

Sum or Difference of Two Cubes

We shall use the following identities:

1. \({a^3} + {b^3} = (a + b)\left( {{a^2} – ab + {b^2}} \right)\)
2. \({a^3} – {b^3} = (a – b)\left( {{a^2} + ab + {b^2}} \right)\)

Let us solve an example to understand better the concept:

Example1: Resolve the algebraic expression \(27{x^3} – \frac{{343}}{{{x^3}}}\) into factors.

Solution: \(27{x^3} – \frac{{343}}{{{x^3}}} = {(3x)^3} – {\left( {\frac{7}{x}} \right)^3}\)
\( = \left( {3x – \frac{7}{x}} \right)\left[ {{{(3x)}^2} + 3x \times \frac{7}{x} + {{\left( {\frac{7}{x}} \right)}^2}} \right]\)
\( = \left( {3x – \frac{7}{x}} \right)\left[ {9{x^2} + 21 + \frac{{49}}{{{x^2}}}} \right]\)

factorisation of polynomial: Solved Examples

Q.1. Factorise the following expression.
\(12ax – 3ay + 8bx – 2by\)
Ans: \(12ax – 3ay + 8bx – 2by = (12ax – 3ay) + (8bx – 2by)\)
\( = 3a(4x – y) + 2b(4x – y)\)
\( = (3a + 2b)(4x – y)\)
Hence, the required factors of the given polynomial is \((3a + 2b)(4x – y)\).

Q.2. Factorise the following polynomial.
\({x^2} + 10x + 25\)
Ans: Observe that \({x^2} = {(x)^2}\) and \(25 = {5^2}\). We would see whether the middle term \(10x\) can be expressed as \(2(x)(5)\). If so, we can apply the perfect square factorisation.
\({x^2} + 10x + 25 = {(x)^2} + 2(x)(5) + {5^2}\)
Apply \({a^2} + 2ab + {b^2} = {(a + b)^2}\)
\( = {(x)^2} + 2(x)(5) + {5^2}\)
\( = {(x + 5)^2}\)
Hence, \({x^2} + 10x + 25\) can be factorized as \((x + 5)(x + 5)\).

Q.3. Evaluate \({238^2} – {237^2}\)
Ans: Apply \({a^2} – {b^2} = (a – b)(a + b)\) by taking \(a = 238\) and \(b = 237\)
\({238^2} – {237^2} = (238 + 237)(238 – 237)\)
\( = 475 \times 1\)
\( = 475\)
Hence, the required answer is \(475\).

Q.4. Factorise the following
\({a^7} – a{b^6}\)
Ans: \({a^7} – a{b^6} = a\left( {{a^6} – {b^6}} \right)\)
\( = a\left( {{{\left( {{a^3}} \right)}^2} – {{\left( {{b^3}} \right)}^2}} \right) = a\left( {{a^3} + {b^3}} \right)\left( {{a^3} – {b^3}} \right)\)
\( = a(a + b)\left( {{a^2} – ab + {b^2}} \right)(a – b)\left( {{a^2} + ab + {b^2}} \right)\)

Q.5. Factorise the following polynomial.
\(16ax – 4ay – 8bx + 2by\)
Ans: \(16ax – 4ay – 8bx + 2by = (16ax – 4ay) – (8bx – 2by)\)
\( = 4a(4x – y) – 2b(4x – y)\)
\( = (4x – y)(4a – 2b)\)
\( = 2(4x – y)(2a – b)\)
Hence, the required factors of the given polynomial are \(2(4x – y)(2a – b)\).

Summary

In this article, we learned about the factors, the definition of polynomials, and a detailed study about the various methods to factorise a polynomial. Various factorisation methods are by grouping the like terms, splitting the middle term, factorisation by regrouping or using the identities.

FAQs

Q.1. How do you factorise polynomials?
Ans: We factorise the polynomials by collecting the like terms together and finding the common factors. It is done by any of the following methods.
(i) Factorisation using common factors
(ii) Factorisation by regrouping
(iii) Factorisation using identities
(iv) Factorisation by splitting the middle term

Q.2. What are the \(4\) types of factoring?
Ans: The \(4\) types of the factorisation are:
(i) Factorisation using common factors
(ii) Factorisation by regrouping
(iii) Factorisation using identities
(iv) Factorisation by splitting the middle term

Q.3. What is the first thing to do when factoring?
Ans: The first step is to group the like terms to take out the common term easily.

Q.4. Define factorisation of polynomials.
Ans: When an algebraic expression can be written as a product of two or more algebraic expressions, each expression is called a factor of the given expression. Thus, finding two or more factors whose product is the given polynomial is called factorisation of polynomials.

Q.5. What is the rule to factorise the trinomial \(a{x^2} + bx + c\), where \(a,\,b\) and \(c\)are real numbers?
Ans: Split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\) and their product is \(ac\), then factorise by grouping method.

We hope this detailed article on the factorisation of polynomials proves to be helpful to you. If you have any doubts or queries regarding this topic, feel to ask us in the comment section.